Answer
(a)
Increasing on $(-2,2)$.
Decreasing on $(-2\sqrt{2}, -2)\cup (2,2\sqrt{2})$
(b)
Absolute maximum: $(2,4)$.
Local maximum. $(-2\sqrt{2},0)$
Absolute minimum: $(-2,-4)$.
Local minimum $(2\sqrt{2},0)$
Work Step by Step
$g$ is defined on $[-2\sqrt{2},2\sqrt{2})$.
$g'(x)=(8-x^{2})^{1/2}+x(\displaystyle \frac{1}{2})(8-x^{2})^{-1/2}(-2x)$
$g'(x)=\displaystyle \frac{2(2-x)(2+x)}{\sqrt{(2\sqrt{2}-x)(2\sqrt{2}+x)}}$
which is not defined at $ x=\pm 2\sqrt{2}\qquad$ ... critical points.
$g'(0)=0$ for $x=\pm 2 \quad$... critical points.
$g(-2)=-4,\qquad g(2)=4,\qquad g(\pm 2\sqrt{2})=0$
Using testpoints in the intervals between critical points,
$g'(-2.1)=-0.4327... \lt 0$
$g'(1)=2.26... \gt 0$
$g'(2.1)=-0.4327... \lt 0$
Create a table using the above:
$\begin{array}{l}
g':\\
\\
\\
g:
\end{array} \begin{array}{lllllll}
-2\sqrt{2} & & -2 & & 2 & & 2\sqrt{2}\\
{[} & -- & & ++ & | & -- & ]\\
& & & & & & \\
0 & \searrow & & \nearrow & 4 & \searrow & \\
& & -4 & & & & 0\\
& & & & & &
\end{array}$
